c++ - 为什么 is_constructible 声称某些东西不是可构造的？

Question

以下程序在使用 GCC 4.7 和 clang 3.2 编译时，会产生“1”作为输出。

#include <type_traits>

struct foo {
    template<typename T>
    foo(T) {
        static_assert(not std::is_same<int, T>(), "no ints please");
    }
};

#include <iostream>    
int main() {
    std::cout << std::is_constructible<foo, int>();
}

这令人困惑。 foo 显然不能从 int 构造!如果我将 main 更改为以下内容，两个编译器都会因为静态断言失败而拒绝它:

int main() {
    foo(0);
}

为什么两个编译器都说它是可构造的？

Answer 1

这是标准必须说的(§20.9.5/6)，我强调:

Given the following function prototype:

template <class T>
typename add_rvalue_reference<T>::type create();

the predicate condition for a template specialization is_constructible<T, Args...> shall be satisfied if and only if the following variable definition would be well-formed for some invented variable t:

T t(create<Args>()...);

[ Note: These tokens are never interpreted as a function declaration. —end note ]

Access checking is performed as if in a context unrelated to T and any of the Args. Only the validity of the immediate context of the variable initialization is considered. [ Note: The evaluation of the initialization can result in side effects such as the instantiation of class template specializations and function template specializations, the generation of implicitly-defined functions, and so on. Such side effects are not in the “immediate context” and can result in the program being ill-formed. —end note ]

只有在模板构造函数被实例化时，断言才会失败。但是，正如注释中所澄清的那样，该断言不在所考虑的变量定义的直接上下文中，因此不会影响其“有效性”。因此编译器可以将该定义视为有效，从而声称 foo确实可以从 int 构造，即使实际上试图构造一个 foo来自 int导致程序格式错误。

请注意，编译器也可以，而不是 is_constructible yield false，只是根据断言拒绝原始程序，即使两者都没有。